# Energy of Activation Clarification

1. Apr 19, 2014

### david_w

In transition state theory, the energy of activation (E) shows up in the Arrenhius equation as: k=Ae^(-E/RT). Does this energy of activation term correspond to internal energy? Or is it some other type of energy (i.e. Gibbs, Helmholtz)? I know its not Gibbs because that appears in the Eyring equation, but I just wanted to clarify that it's internal energy.

2. Apr 19, 2014

### Staff: Mentor

Transition state theory deals with kinetics, Gibbs and Helmholtz energies deal with thermodynamics. Mixing these things never helps.

3. Apr 19, 2014

### david_w

Yes, but the thermodynamics of the transition state affects the kinetics.

4. Apr 20, 2014

### DrDu

David_w is right in that in transition state theory, the rate depends on the Gibbs free energy of the activated complex

Arrhenius equation is purely empirical. You can use $E=-d ln k/d((RT)^{-1})$ to define E, using the expression for k from transition state theory. In this derivation, you have to take in mind that $\Delta G$ also depends on T.
http://de.wikipedia.org/wiki/Eyring-Theorie

Last edited: Apr 20, 2014
5. Apr 20, 2014

### david_w

Here's my understanding.

The activation energy is defined as the minimum amount of energy required to bring about formation of the transition state. This can be confirmed by the use of potential-energy surfaces like this. This plots potential versus different states of this system. In traveling along the minimum path from reactants to products, the highest point (the saddle point) represents the highest potential energy of the system. This height of this relative to the reactants represents the activation energy. Since the work done by a conservative force is equal to the negative change in the potential energy, the activation energy does indeed represent the minimum energy requirement--all of this is consistent.

Let's turn now to Transition-State Theory and the Eyring equation. Using the Eyring equation in the form: $k=\frac{k_B T}{h} K^\ddagger$ along with the Gibbs-Helmholtz equation (as you have mentioned): $\frac{d ln{K^\ddagger}}{dT}=\frac{∆^\ddagger U°}{R T^2}$, you get the following relation: $E_a=RT + ∆^\ddagger U°$.

The last equation says that energy of activation is not internal energy then? I'm not sure what it is then. From an explanation of the Arrhenius equation I've found, it certainly looks like internal energy. This explanation is as follows: $\frac{d ln{K_c°}}{dT}=\frac{∆U°}{R T^2}$. The equilibrium constant is the ratio of the rate constants: $K_c=\frac{k_1}{k_{-1}}$ and the internal energy change is equal to the difference between the activation energy for the forward reaction and the reverse reaction: $∆U°=E_1-E_{-1}$. Plugging all of this into the Gibbs-Helmholtz equation and solving yields: $k_1=A_1 e^{-E_1/RT}$ where A is the pre exponential factor and is determined empirically.

Here is where I am really confused. In the second paragraph, we relate activation energy and the change in internal energy. In the third paragraph, it looks like activation energy IS internal energy. This all seems very inconsistent to me. This is all coming from the same source btw (Physical Chemistry by Laidler, Meiser, and Sanctuary).